Homework Help: Triangle problem

1. Dec 23, 2007

ehrenfest

1. The problem statement, all variables and given/known data
A point P is located in the interior of an equilateral triangle ABC. Perpendiculars drawn from P meet each of the sides in points D,E,F, respectively. Where should P be located to make PD + PE + PF a maximum? What about a minimum?

2. Relevant equations

3. The attempt at a solution

From symmetry considerations. I think that the maximum will be at the incenter. I am not sure about the minimum. Maybe at a vertex, but that is not really in the interior... I tried the standard reflections. I am not sure if I actually need to set up a coordinate system with P at the origin... That looks like it could get messy? I am also not sure how to use the fact that this is an equilateral triangle.

2. Dec 23, 2007

ehrenfest

anyone?

3. Dec 24, 2007

marlon

Isn't this one of them Lagrange multiplier things ?

Having said that, i don't know how to set up :

1)the funtion that needs to be minimized/maximized

2) the function that is going to be the constraint.

I would reallly like to see the solution to this problem, though.

Marlon

4. Dec 24, 2007

jacobrhcp

I think it's my english that's bad here, but how can you draw 3 perpendicular lines in a plane?

5. Dec 24, 2007

chickendude

They are not perpendicular to each other.

They are perpendicular to each side of the triangle.

6. Dec 24, 2007

chickendude

Draw segments from P to each vertex of the triangle. You have split the triangle into three new triangles.

The area of each triangle is given by $$\frac{1}{2}sh_i$$ where s is the side length of the equilateral triangle and h_i is the height of the perpendicular on that side

Through this, we can set up the equality

$$\frac{1}{2}sh_a + \frac{1}{2}sh_b + \frac{1}{2}sh_c = \frac{s^2 \sqrt{3}}{4}$$

$$h_a + h_b + h_c = \frac{s \sqrt{3}}{2}$$

meaning that the sum of the three perpendiculars is always constant regardless of the location of P

7. Dec 24, 2007

jacobrhcp

well, take the situation that P is at point a, then if you move D a distance d, for instance, you can still hold E still, but you'll have to move F to keep P one point.

now P moves in the direction of E with a distance: 2d/root[3], this is the distance it shrinks.
the distance it grows is: 2d tan(30) = 2d/root[3], this is equal.

because any motion of P can be described as the sum of the motion in the direction of perpendiculars, the distance is always the same.

Last edited: Dec 24, 2007
8. Dec 24, 2007

jacobrhcp

I must have done something wrong, because chickendude seems right.

EDIT: I've edited my post and saw what i did wrong.

Last edited: Dec 24, 2007
9. Dec 24, 2007

Rainbow Child

10. Dec 24, 2007

ehrenfest

Hmmm. I did not see anything wrong with chickendude's post, which means there is probably something wrong here.

11. Dec 24, 2007

NateTG

If you consider the point $p$ to be the tip of three triangles, each of which has one of the sides of the original equilateral triangles as it's base then the total area of those triangles will be:
$$\frac{1}{2} l_1 s + \frac{1}{2} l_2 s + \frac{1}{2} l_3 s = \frac{1}{2} (l_1 + l_2 +l_3) s$$
which is also the area of the original triangle, which is clearly constant, so
$$(l_1+l_2+l_3)$$
must be constant.

12. Dec 24, 2007

ehrenfest

Yes. You just repeated chickendude's proof. I am just saying if that is right, then something is wrong with Rainbow Child's proof.

13. Dec 24, 2007

Rainbow Child

My mistake!

In my proof I took the points

$$E(\frac{\alpha}{4},\frac{\sqrt{3}\,\alpha}{4}),\,F (\frac{3\,\alpha}{4},\frac{\sqrt{3}\,\alpha}{4}),\ ,D(\frac{\alpha}{2},0)$$

which are the midpoints of the sides AB, AC and AD! (nothing to do with the original problem )

So I gave the answer to the question:

"Which point extemizes the sum $$|PD|+|PE|+|PF|$$ when D, E and F are the midpoints of the sides!"

Sorry!